The locus of the midpoints of the intercepted | vedclass

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(A) The given equation of the ellipse is $x^2+2y^2=2$,which can be written as $\frac{x^2}{2}+\frac{y^2}{1}=1$. Here $a^2=2$ and $b^2=1$,so $a=\sqrt{2}

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$\frac{1}{2x^2}+\frac{1}{4y^2}=1$

Vedclass · Answer

(A) The given equation of the ellipse is $x^2+2y^2=2$,which can be written as $\frac{x^2}{2}+\frac{y^2}{1}=1$. Here $a^2=2$ and $b^2=1$,so $a=\sqrt{2}$ and $b=1$.The equation of the tangent to the ellipse at point $P(a \cos 	heta, b \sin 	heta)$ is $\frac{x \cos 	heta}{a} + \frac{y \sin 	heta}{b} = 1$.Substituting the values,we get $\frac{x \cos 	heta}{\sqrt{2}} + y \sin 	heta = 1$.For the $x$-intercept $(A)$,set $y=0$: $\frac{x \cos 	heta}{\sqrt{2}} = 1 \Rightarrow x = \sqrt{2} \sec 	heta$. So,$A = (\sqrt{2} \sec 	heta, 0)$.For the $y$-intercept $(B)$,set $x=0$: $y \sin 	heta = 1 \Rightarrow y = \operatorname{cosec} 	heta$. So,$B = (0, \operatorname{cosec} 	heta)$.Let $M(h, k)$ be the midpoint of $AB$. Then:$h = \frac{\sqrt{2} \sec 	heta + 0}{2} = \frac{\sec 	heta}{\sqrt{2}} \Rightarrow \sec 	heta = \sqrt{2}h$$k = \frac{0 + \operatorname{cosec} 	heta}{2} = \frac{\operatorname{cosec} 	heta}{2} \Rightarrow \operatorname{cosec} 	heta = 2k$We know that $\cos^2 	heta + \sin^2 	heta = 1$,which implies $\frac{1}{\sec^2 	heta} + \frac{1}{\operatorname{cosec}^2 	heta} = 1$.Substituting the values of $\sec 	heta$ and $\operatorname{cosec} 	heta$:$\frac{1}{(\sqrt{2}h)^2} + \frac{1}{(2k)^2} = 1$$\frac{1}{2h^2} + \frac{1}{4k^2} = 1$Replacing $(h, k)$ with $(x, y)$,the locus is $\frac{1}{2x^2} + \frac{1}{4y^2} = 1$.

The locus of the midpoints of the intercepted portion of the tangents by the coordinate axes,which are drawn to the ellipse $x^2+2y^2=2$,is

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